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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 351. |
Assertion : Hydrochloric acid is not used to acidify a `KMnO_(4)` solution in volumetric analysis of `Fe^(2+)` and `C_(2) O_(4)^(2-)` because. Reason : Part of the oxygen produced from `KMnO_(4)` and `HCl` is used up in oxidising `HCl` to `Cl_(2)`.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Assertion and reason both are correct and reason is the correct explanation of assertion. `2HC1 + [O] rarr H_(2) O + C1_(2) (s)` |
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| 352. |
Correct statement about FeO at room temperatureA. It is non-stoichiometirc and metal deficientB. It is basic oxideC. Its aqueous solution changes to `Fe(OH)_(3)` and then to `Fe_(2)O_(3),xH_(2)O` by atmospheric oxygenD. It gives red colour with KCNS |
| Answer» Correct Answer - A::B::C | |
| 353. |
Which one of the transittion metal ions is coloured in aqueous solution?A. `Cu^(+)`B. `Zn^(2+)`C. `Sc^(3+)`D. `V^(4+)` |
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Answer» Correct Answer - D (a) Valence shell electron configuration of `Cu^( +)` is `[Ar]^(18) 3 d^(10)`, so `n = 10` (b) Valence shell electron configuration of `Zn^( 2+)` is `[Ar]^(18) 3 d^(10)`, so `n = 0` (c ) Valence shell electron configuration of `Sc^(3 +)` is `[Ar]^(18) 3 d^(0)`, so `n = 0` (d) Valence shell electron configuration of `V^( 4+)` is `[Ar]^(18) 3 d^(1)`, so `n = 1` As `V^(4+)` has one uparied electron and so in presence of water as ligand it will undergo d-d transition of electron Hence `V^(4 +)` ions will produce colure in the solution. |
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| 354. |
The `(+5)` oxidation state is by far the most stable and best known forA. VB. NbC. TaD. Ag Nb and Ta |
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Answer» Correct Answer - D Group 5 consists of elements- V, Nb, Ta and Db. In many ways, these elements are similar to the elements of Group 4. They are shiny silver coloured metals with high melting points. Keeping with the trend, they are slightly less positive than their predeccessor. Although each of these elements shows all the formal oxidation states form `(+5)` down to `(-1)`, the most stable one in case of V is `+4`. By contrast, the chemistry of Nb and Ta is confined to the group oxidation state of `+5`. Element V readily exists in four different oxidation states: `+5, +4, +3`, and `+2`, corresponding to the `d^(0)` , `d^(1)` , `d^(2)` , and `d^(3)` electron configurations. The chief ue of V is to increase the strength and toughness of steel. |
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| 355. |
Which of the following pairs is coloured in aqueous solution?A. `Sc^(3+),Co^(3+)`B. `Ni^(2+),Cu^(2+)`C. `Ni^(2+),Ti^(3+)`D. `Sc^(3+),Ti^(3+)` |
| Answer» Correct Answer - B::C | |
| 356. |
In which of the following pairs do the elements have nearly the same densities ?A. Cr and FeB. Zn and MnC. Co and NiD. Ag and Au |
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Answer» Correct Answer - C `Cr(7.14g cm^(-3))`, `Mn(7.43g cm^(-3))`, `Fe(7.87g cm^(-3))`, `Co(8.90g cm^(-3))`, `Ni(8.91g cm^(-3))`, `Ag(10.49g cm^(-3))` and `Au(19.32)`. |
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| 357. |
Assertion : `4d` and `5d` series elements have nearly same atomic radius. Reason : Lanthanoid contraction.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A The atomic radii of the second (`4d` series ) and third, (`5d` series) transition series are almost the same. This phenomenon is associated with te intervention of the `4f` orbitals which must be filled before the `5d` series of element begin. The filling of `4f` before `5d` orbital results in a regular decrease in atomic radii called Lanthanid contraction which essentially compensates fro the expected in atomic size with incresing atomic number. The net result of the lanthanoid contraction is the the second and the third `d` series exhibit similar radii (e.g., `Zr 160 "pm", Hf 159 "pm")`. |
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| 358. |
Assertion : Solution of `Na_(2) CrO_(4)` in water is intensely electrons. Reason : Oxidation state of `Cr` in `Na_(2) CrO_(4)` is `+ VI`.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A A solution of `Na_(2)CrO_(4)` in water is intensely coloured due to oxidation state of chromium in `Na_(2) CrO_(4)` is `+ 6`. Here both assertion and reason are corredt. |
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| 359. |
STATEMENT-1 : `CrO_(2)Cl_(2)` has tetrahedral shape . and STATEMENT-2: `CrO_(3)` reacts with HCl to form ` CrO_(2)Cl_(2)` .A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - B | |
| 360. |
Because of lanthnoid contractoin, which of the following pairs of elements have nearly same atomic radii ? (Number in the parentrhesis are atomic numbers)A. `Zr (40)` and `Ta (73)`B. `Ti (22)` and `Zr (40)`C. `Zr (40` and `Nb (41)`D. `Zr (40)` and `Hf (72)` |
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Answer» Correct Answer - D As a result of lanthanoid contraction the normal increase in size form `Sc rarr Y rarr La` disapperas after the lanthanoids, and thus pairs of elements such as Zr and Hf (Group 4), Nb and Ta (Group 5), Mo and W (Group 6), etc., possess nearly similar sizes. Consequently, the properties of these elements are very similar. The similarites in properties make their separtion very difficult. |
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| 361. |
The number of `d` electrons in `Fe^(2+)` (atomic number of `Fe = 26`) is not equal to that of the.A. p- electrons in Ne (Z = 10)B. s- electrons Mg (Z = 12)C. p- electrons in CI (Z = 17)D. d- electrtons in Fe (Z = 26) |
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Answer» Correct Answer - C `Fe(3d^(6) 4s^(2)) rarr Fe^(2+) (3d^(6))` `Ne (1s^(2) 2s^(2) 2p^(6)), Mg (1s^(2) 2s^(2) 2p^(6) 3s^(2))` and `CI(1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(5))` Thus, Fe has six 3d electrons, `Fe^(2+)` has six 3d electrons, Ne has six 2p electrons, Mg has six s- electrons and CI has eleven p elements. |
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| 362. |
`4K_(2) Cr_(2) O_(7) overset(heat)rarr 4K_(2) CrO_(4) + 3O_(2) + X`. In the above reaction `X` isA. `CrO_(3)`B. `Cr_(2)O_(7)`C. `Cr_(2)O_(3)`D. `CrO_(5)` |
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Answer» Correct Answer - C Potassium dichromate, on heating gives oxygen and chromic oxide `(Cr_(2)O_(3))`. `4K_(2) Cr_(2) O_(7) overset(Delta)rarr 4K_(2) Cr_(2) O_(4) + 3O_(2) + 2Cr_(2) O_(3)` |
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| 363. |
Assertion : `CrO_(3)` is an acid anhydride. Reason: `CrO_(3)` is obtained as bright orange crystals by the reaction of `K_(2) Cr_(2) O_(7)` with cold concentrated `H_(2) SO_(4)`.A. If both assetion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are ture but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B Assertion: Correct statement as it gives oxy acid `H_(2) Cr_(2) O_(7)`, `2CrO_(3) + H_(2)O rarr H_(2) Cr_(2) O_(7)`, Reason : Correct statement: `K_(2) Cr_(2) O_(7) + 2H_(2) SO_(4)` (connected and cold) `rarr CrO_(3)` (bright orange) `+ 2KHSO_(4) + H_(2)O`. |
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| 364. |
The reaction of aqueus `KMnO_(4)` with `H_(2)O_(2)` in acidic conditions givesA. `Mn^(4+)` and `O_(2)`B. `Mn^(2+)` and `O_(2)`C. `Mn^(2+)` and `O_(3)`D. `Mn^(4+)` and `MnO_(2)` |
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Answer» Correct Answer - B In acidic medium `KMnO_(4)` is reduced to manganaous ion `(Mn^(2+))` while `H_(2)O_(2)` is oxidized to `O_(2)` : `{:(2KMnO_(4)+3H_(2)SO_(4)rarrK_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[O]),(" "H_(2)O_(2)+[O]rarrH_(2)O+[O_(2)]xx5),(bar(2KMnO_(4)+3H_(2)SO_(4)+5H_(2)O_(2)rarrK_(2)SO_(4)+2MnSO_(4)+H_(2)O+5O_(2))):}` or `2MnO_(4)^(-) + 5H_(2)O_(2) + 6H^(+) rarr 2Mn^(2+) + 8H_(2)O + 5O_(2)` |
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| 365. |
In which of the following compounds transition metal has zero oxidation stae ?A. `[Fe (CO)_(5)]`B. `NH_(2) . NH_(2)`C. `NOCIO_(4)`D. `CrO_(5)` |
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Answer» Correct Answer - A In metal carbonyls such as `[Fe(CO)_(5)]`, oxidation state of metal is always zero because the carbonyl is a netural species and carbon monoxide is a netural ligand. |
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| 366. |
Within each transition series, the oxidation statesA. first increase till the middle of the period and then decreaseB. decrease regularly in moving form left to rightC. first increase till the middle of period and then increaseD. increase regularly in moving from left to right |
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Answer» Correct Answer - A There exists a general trend of lesser number of oxidation states at each end of the series and a higher number in the middle. The lesser number of oxidation states in the beginning of the series is due to the presence of too few (n - 1) d electrons to lose or share, towards the end of the series it can be ascribed to the presence of too many (n - 1) d electrons and thus fewer unpaired electrons as well as empty orbitals to share electrons with the ligands. |
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| 367. |
While extracting an element from its ore, the ore is grind and leached with dil KCN solution to form the soluble product potassium argentocyanide. The element isA. MnB. CrC. FeD. Ag |
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Answer» Correct Answer - D Silver is obtained form its low grade ores by leaching them whith a dulute solution of sodium cyanide, when the silver compounds dissolve forming the soluble complex argentocyanide ion: `Ag_(2)S + 4 NaCNhArr2Na [Ag (CN)_(2)] + Na_(2)S` Air is blown into the mixture to oxidize the sodium sulphide to sulphate and thereby make the reversible reaction proceed completely to the right. `4 Na_(2)S + 5O_(2) + 2H_(2)O rarr 2 Na_(2)SO_(4) + NaOH + 2S` The silver is precipitated form the solution by adding some electropositive metal, usually Zn or AI: `2Na [Ag(CN)_(2)] + Zn rarr 2 Ag + Na_(2) [Zn(CN)_(4)]` |
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| 368. |
Among the following series of transition metal ions the one where all meal ions have `3d^2` electronic configuration isA. `Ti^(3+),V^(2+),Cr^(3+),Mn^(4+)`B. `Ti^(+),V^(4+),Cr^(6+),Mn^(7+)`C. `Ti^(4+),V^(3+),Cr^(2+),Mn^(3+)`D. `Ti^(2+),V^(3+),Cr^(4+),Mn^(5+)` |
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Answer» Correct Answer - D `Ti^(2+) = 1 s^(2) 2 s^(2) 2 p^(6) 3 s^(2) 3 p^(6) 3 d^(2) 4 s^(0)` `V^(3+) = 1 s^(2) 2 s^(2) 2 p^(6) 3 s^(2) 3 p^(6) 3 d^(2) 4 s^(0)` `Cr^(4+) = 1 s^(2) 2 s^(2) 2 p^(6) 3 s^(2) 3 p^(6) 3 d^(2) 4 s^(0)` `Mn^(5+) = 1 s^(2) 2 s^(2) 2 p^(6) 3 s^(2) 3 p^(6) 3 d^(2) 4 s^(0)` |
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| 369. |
Among `K,Ca,Fe` and `Zn` the element which can form more than one binary compound with chlorine isA. FeB. ZnC. KD. Ca |
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Answer» Correct Answer - A Of the given elements only iron shows variable oxidation states. Its most important oxidation states are `+2 (FeCI_(2))` and `+3 (FeCI_(3))` . |
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| 370. |
Among the following series of transition metal ions the one where all meal ions have `3d^2` electronic configuration isA. `Ti^(3+) , V^(2+) , Cr^(3+) , Mn^(4+)`B. `Ti^(+)` , `V^(4+)` , `Cr^(6+)` , `Mn^(7+)`C. `Ti^(+)` , `V^(4+)` , `Cr^(6+)` , `Mn^(7+)`D. `Ti^(2+)` , `V^(3+)` , `Cr^(4+)` , `Mn^(5+)` |
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Answer» Correct Answer - D `Ti (3d^(2) 4s^(2)) implies Ti^(+) (3d^(2) 4s^(1)) implies Ti^(2+) (3d^(2)) implies Ti^(3+) (3d^(1))` `V (3d^(3) 4s^(2)) implies V^(2+) (3d^(2)) implies V^(4+) (3d^(1))` `Cr (3d^(5) 4s^(1)) implies Cr^(3+) (3d^(3)) implies Cr^(4+) (3d^(2)) implies Cr^(6+) (3d^(0))` `Mn (3d^(5) 4s^(2)) implies Mn^(4+) (3d^(3)) implies Mn^(5+) (3d^(2)) implies Mn^(7+) (3d^(0))` |
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| 371. |
Which one of the following characteristics of transition metals is associated with their catalytic activity?A. High enthalpy of atomizationB. Paramagnetic behaviourC. Colour of hydrated ionsD. Variable oxidation states |
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Answer» Correct Answer - D Compounds of transition metals are good catalysts on account of variable oxidation states exhibited by transition metals. |
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| 372. |
STATEMENT-1 : In `Cr_(2)O_(7)^(-2)` bond length of all Cr-O bond is equal and STATEMENT-2: `Cr_(2)O_(7)^(-2)` , resonance is possible.A. Statement-1 is True , Statement-2 is True , Statement-2 is a correct explanation for Statement-1B. Statement-1 is True , Statement-2 is True , Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True , Statement-2 is FalseD. Statement-1 is False , Statement-2 is True |
| Answer» Correct Answer - D | |
| 373. |
`H_(2)Cr_(2)O_(7)` on heating with aqueous `NaOH` givesA. `Cr_(4)^(2-)`B. `Cr(OH)_(3)`C. `Cr_(2)O_(4)^(2-)`D. `Cr(OH)_(2)` |
| Answer» Correct Answer - A | |
| 374. |
`KMnO_(4)` react with oxalic acid according to the equation, `2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+) rarr 2Mn^(2+)+ 10 CO_(2)+8H_(2)O`, here `20 ml` of `0.1 M KMnO_(4)` is equivalemt toA. `20 ml of0.5 M C_(2)H_(2)O_(4)`B. `50 ml of 0.1 M C_(2)H_(2)O_(4)`C. `20 ml of 0.5 M C_(2)H_(2)O_(4)`D. `20 ml of 0.1 M C_(2)H_(2)O_(4)` |
| Answer» Correct Answer - B | |
| 375. |
The interstitial compounds of transition metals are than the metal itself.A. softerB. more malleableC. harderD. more metallic |
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Answer» Correct Answer - C Interstitial hydrides, borides, carbides and nitrides are ill-defined compounds formed by a number of transition metals in which small atoms H, B, C and N respectively are accommodated in the voids or interstices in the crystal lattice of the transition element. On account of the filling of vacant spaces and formation of extra bonds, interstitial compounds have following proerties: (i) They are chemically insert (ii) They retrain metallic conductivity (iii) They are very hard. Some borides approach diamond in hardness. (iv) They have high melting points, higher than those of pure metals |
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| 376. |
Incorrect statement isA. In acidic medium `MnO_(4)^(-)` disproportionates to `MnO_(2)` and `MnO_(4)^(2-)`B. `KMnO_(4)` spot can be bleached by `H_(2)O_(2)`C. Alkaline `KMnO_(4)` can be used to test unsaturation is D. Eq. wt. of `KMnO_(4)` in acidic medium is `(M)/(5)` |
| Answer» Correct Answer - A::C | |
| 377. |
In context with the transition elements, which of the following statements is incorrect ?A. In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes.B. In the highest oxidation state, the transition metals show basic character and form cationic complexes.C. In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding.D. Once the `d^5` configuration is exceeded, the tendency to involve all the 3d- electrons in bonding decreases. |
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Answer» Correct Answer - B When the trasition metals are in their highest oxidation state, they no longer have tendency to give away electrons , thus they are not basic but show acidic character and from anionic complexes. |
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| 378. |
Cerium `(Z= 58)` is an important nember of the lanthanoids . Which of the following statements about cerium is incorrect ?A. The common oxidation states of cerium are +3 and +4.B. The +3 oxidation state of cerium is more stable tan +4 oxidation state.C. The +4 oxidation state of cerium is not known in solutions.D. Cerium (IV)acts as an oxidising agent. |
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Answer» Correct Answer - C `+4` oxidation state of cerium is also known in solutions. |
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| 379. |
Identify the incorrect statement among the following:A. As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements.B. Shielding power of 4d electrons is quite weakC. There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.D. Lanthanoid contraction is the accumulation of successive shrinkages |
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Answer» Correct Answer - A Elements of 4f series and 5d series show similarities in properties because the size of elements of 4d series are almost similar to those of 5d series on account of lanthanoid contraction which is due to the greater effect of the increased nuclear charge than of screening effect (shielding effect). |
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| 380. |
Which of the following statements about the interstitial compounds is incorrect?A. They are chemically reactiveB. They much harder than the pure metalC. They have higher melting points than the pure metalD. They retain metallic conductivity |
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Answer» Correct Answer - A They are chemically inert because all the (n-1) d electrons are utilized to form bonds with the atoms of nonmetals which are trapped within the voids or interstices of crystal lattice. |
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| 381. |
Identify the incorrect statement among the following.A. 4f and 5f orbitals are equally shielded.B. d-Block elements show irregular and erratic chemical properties among themselves.C. La and Lu have partially filled d-orbitals and no other partially filled orbitals.D. The chemistry of various lanthanoids is very similar. |
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Answer» Correct Answer - A The decrease in the force of attraction exerted by the nucleus on the valence electrons due to presence of electrons in the inner shells is called shielding effect. A 4f orbital is nearer to the nucleus than 5f orbital. Hence shielding of 4f is more than 5f. |
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| 382. |
The correct of decreasing second ionisation enthalpy of `Ti(22),V(23),Cr(24)` and `Mn(25)` isA. `Ti gt V gt Cr gt Mn`B. `Cr gt Mn gt V gt Ti`C. `V gt Mn gt Cr gt Ti`D. `Mn gt Cr gt TI gt V` |
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Answer» Correct Answer - B In general, ionization enthalpy increases as we move across the period on account of increases of effective nuclear charge. But Cr has unsually high `Delta_(i)H_(2)` on account of extra stability of `d^(5)` configuration of `Cr^(+)` ion. Thus, the correct order is `Cr gt Mn gt V gt Ti` |
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| 383. |
`MnF_(7)` is not stable while `Mn_(2)O_(7)` is stable. |
| Answer» `MnF_(7)` is unstable due to more crowd of F which lead to increase in repulsion | |
| 384. |
Acidified `K_2Cr_2O_7` solution turns green when sodium sulphite is added to it. Explain.A. `CrSO_(4)`B. `Cr_(2)(SO_(4))_(3)`C. `CrO_(4)^(2-)`D. `Cr_(2)(SO_(3))_(3)` |
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Answer» Correct Answer - B Sulphites are oxidized to sulphates: `{:(K_(2)Cr_(2)O_(7) + 4H_(2)SO_(4) rarr K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 4H_(2)O + 3[O]),(" "[Na_(2)SO_(3) + [O] rarr Na_(2)SO_(4)]xx3),(bar(K_(2)Cr_(2)O_(7) + 3Na_(2)SO_(3) + 4H_(2)SO_(4) rarr 3Na_(2)SO_(4) + K_(2)SO_(4) + Cr_(2)(SO_(4))_(3) + 4H_(2)O)):}` or `Cr_(2)O_(7)^(2-) + 3SO_(3)^(2-) + 8H^(+) rarr 3SO_(4)^(2-) + 2Cr^(3+) + 4H_(2)O` |
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| 385. |
Write the equations . (i) Acidified solution of `K_(2)Cr_(2)O_(7)` turns green when sodium sulphite is added to it. (ii) Potassium ferricyanide is added to ferrous sulphate . (iii) Silver chloride is treated with aqueous sodium cyanide . (iv) Zinc is exposed to moist air . |
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Answer» (i) `Cr_(2)O_(7)^(-2)+3SO_(3)^(-2)+8H^(+) to 2Cr^(+3)+3SO_(4)^(-2)+4H_(2)O` (ii) `K_(3)[Fe(CN)_(6)]+FeSO_(4) to Kfe[Fe(CN)_(6)]+K_(2)SO_(4)` (iii) ` AgCl+2NaCN to Na[Ag(CN)_(2)]+NaCl` (iv) ` 4Zn+2O_(2)+3H_(2)O+CO_(2) to underset("(Basic Zinc carbonate)")(ZnCO_(3).3Zn(OH)_(2))` |
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| 386. |
The correct order of ionic radii of Ce, La, Pm and Yb in +3 oxidation state isA. `La^(3+) lt Pm^(3+) lt Ce^(3+) lt Yb^(3+)`B. `Yb^(3+) lt Pm^(3+) lt Ce^(3+) lt La^(3+)`C. `La^(3+) ltCe^(3+) ltPm^(3+)lt Yb^(3+)`D. `Yb^(3+) lt Ce^(3+) ltPm^(3+) lt La^(3+)` |
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Answer» Correct Answer - B The ocerall decrease in atomic and ionic radii from `La^(3+)" to " Lu^(3+)` is called lanthanoid contraction. Hence, the correct order is `Yb^(3+) lt Pm^(3+) lt Ce^(3+) lt La^(3+)` |
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| 387. |
Ammonium dichromate is used in some fireworks. The green-coloured powder blown in the air isA. `Cr`B. `CrO_(3)`C. `Cr_(2)O_(3)`D. `CrO (O_(2))` |
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Answer» Correct Answer - C `(NH_(4))_(2) Cr_(2) O_(7) overset(Delta)rarr 2K_(2) CrO_(4) + Cr_(2) O_(3) + (3)/(2) O_(2)` |
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| 388. |
When `MnO_2` is fused with KOH and `O_2`, what is the product formed and its colour? `MnO_2+KOH +O_2 to"_________" + H_2O`A. MnO - colourelessB. `K_2MNO_4` - purpleC. `K_2MnO_4` -dark greenD. `MnO_3` -black |
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Answer» Correct Answer - C `2MnO_2+4KOH+O_2 to ("dark green")(2K_2MnO_4)+2H_2O` |
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| 389. |
What is the oxidation state of Mn in product formed by the oxidising action of `KMnO_(4)` inneutral medium ? |
| Answer» Correct Answer - 4 | |
| 390. |
`KMnO_(4)` in basic medium is reduced toA. `K_(2)MnO`B. `Mn(OH)_(2)`C. `MnO_(2)`D. `Mn^(2+)` |
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Answer» Correct Answer - C `KMnO_(4)` is first reduced to manganate and then to insoluble manganese dioxide. Colour changes first from purple to green and finally becomes colourless. `2KMnO_(7) + 2 KOH rarr 2K_(2) MnO_(4) + H_(2) O + O` `2KMnO+(4) + 2 H_(2) O rarr 2 MnO_(4) + 4KOH + 2O` `2KMnO_(2) + H_(2)O overset(a l k a l i n e)rarr 2MnO_(2) + 2KOH + 3 [O]` |
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| 391. |
Potassium dichromate is prepared fromA. chromate obtined by the fusion of chromite ore with sodium carbonate in free access of airB. pyrolusite which is fused with potassium hydroxide in th presence of airC. iron pyrites by the fusion with potassium carbonate in presence of moistureD. none of these |
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Answer» Correct Answer - A `underset("Chromite ore")(4FeCr_2O_4)+8Na_2Xo_3+7O_2tounderset("(yellow)")underset("Sodium chriomate")(8Na_2CrO_4)+2Fe_2O_3+8CO_2` `2Na_2CrO_4+2H^(+)toNa_2Cr_2O_7+2Na^(+)+H_2O` `Na_2Cr_2O_7+2HCI to underset("dichromate")underset("Potassium")(K_2Cr_2O_7)+2NaCl` |
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| 392. |
Arrange the oxides of manganese according to increasing acidic strength.A. `MnO lt Mn_3O_4 lt Mn_2O_3 lt MnO_2 lt Mn_2O_7`B. `Mn_2O_7 lt MnO_2 lt Mn_2O_3 lt Mn_3O_4 lt MnO`C. `MnO_2 lt Mn_2O_7 lt Mn_3O_4 lt Mn_2O_3 lt MnO`D. `Mn_3O_4 lt Mn_2O_3 lt Mn_2O_7 lt MnO_2 lt MnO` |
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Answer» Correct Answer - A Acidic strength of oxides of transition metals increases with increases in oxidation number. `overset(+2)(MnO),overset(+8//3)(Mn_3O_4),overset(+3)(Mn_2O_3),overset(+4)(MnO_2),overset(+7)(Mn_2O_7)` Hence acidic strength is of the order of `underset("Basic")(MnO)lt Mn_2O_4 lt underset("Amphoteric")(Mn_2O_3) lt MnO_2 lt underset("Acidic")(Mn_2O_7)` |
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| 393. |
Across the lanthanide series, the basicity of the lanthanoid hydroxides:A. increaseB. DecreasesC. first increases and then decreasesD. does not change |
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Answer» Correct Answer - B Across lanthanoid series basicity of lanthanoid hydroxide decreases. |
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| 394. |
How many equivalents of oxygen atoms can be available from 1 mole of `K_(2)Cr_(2)O_(7)` ?A. 6B. 3C. 5D. 2 |
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Answer» Correct Answer - B Equivalent mass of `K_(2)Cr_(r)O_(7)` is formula mass/6. Thus, one equivsalent of `K_(2)Cr_(2)O_(7)` contains six equivalents of `K_(2)Cr_(2)O_(7)`, which (according to low of equivalents) will release six equivalents of oxyge. Equivalent mass of oxygen is 8u. Thus, one equivalents of oxygen 8g is equal to half mole of oxygen atoms (atomic mass of oxygen is 16u). This implies that six equivalents of oxygen correspond to `6xx1//2`, i.e., 3 moles of oxygen atoms. |
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| 395. |
The number of moles of `K_(2)Cr_(2)O_(7)` reduced by `1 mol` of `Sn^(2+)` ions isA. 3B. `1//6`C. 6D. `1//3` |
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Answer» Correct Answer - D `{:(Cr_(2)O_(7)^(2-),rarr,Cr^(3+),Sn^(2+),rarr,Sn^(4+)),(Cr(+6),,Cr(+3),Sn(+2),,Sn(+4)):}` One formula unit of dichromate ion contains two Cr atoms. Thus, total change in oxidation number is of 6 units `(+12` to `+6)` . Thus, 1mole of `Cr_(2)O_(7)^(2-)` contains 6 equivalents. For tin, the oxidation number change by 2 units. Thus, 1 mole of `Sn^(2+)` ion contains 2 equivalents. According to law of equivalents, two equivalentsof `Sn^(2+)` ion will only reduce 2 equivalents of `Cr_(2)O_(7)^(2-)`. 6 equivalents of `Cr_(2)O_(7)^(2-)` come form 1 mole 1 equivalent will come form `(1)/(6)` mole 2 equivalents will come form `(2)/(6)` mole i.. `1//3` mole. |
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| 396. |
The equilibrium `Cr_(2)O_(7)^(2-)hArr 2CrO_(4)^(2-)`A. an acidic mediumB. an alkaline mediumC. a netural mediumD. It does not shift. |
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Answer» Correct Answer - A The addition of an acid to an aqueous solution of a chromate ion, `CrO_(4)^(2-)`, results in the elmination of water and the formation of a condensed anion - the dichromate anion `Cr_(2)O_(7)^(2-)`. The colour of the solution changes form yellow (the color of `CrO_(4)^(2-)`) to orange (the color of `Cr_(2)O_(7)^(2-)`). The reaction is easily reversed by adding alkai`(OH^(-)` ions): `underset("Yellow")(2CrO_(4)^(2-)) (aq) + 2H^(+) (aq)hArr(aq) + H_(2)O (1)` |
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| 397. |
Why d block metals show variable oxidation state ? |
| Answer» Variability of oxidation states , a characteristic of transition elements is due to incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity e.g. , `V^(II),V^(III),V^(IV),V^(v)`. | |
| 398. |
How can we show that change in oxidation state is different from variable oxidation state for non transition element ? |
| Answer» The variability of oxidation states of higher oxide of Fe above `Fe_(2)O_(3)` are known . Although Ferrates `(FeO_(4))^(-2)` are formed in alkaline medium but readily decompose to `Fe_(2)O_(3) and O_(2)` . Beside the oxides , oxocations stabilises `V^(V) as VO_(2)^(+), V^(IV) as VO^(2+) and Ti^(IV) as TiO^(2+)`. The ability of oxygen to stabilise these high oxidation state except with fluoride . Thus, the highest Mn fluorides is `MnF_(4)` where as the highest oxide is `Mn_(2)O_(7)`. The ability of oxygen to form multiple bonds wiht metals explain its stronger oxidising action. | |
| 399. |
Complate the following reactions. (i) `Cr_2O_7^(2-)+3SO_2+2H^(+)to2Cr^(3+)+"________"+H_2O` (ii) `2MnO_4^(-)+5SO_3^(2-)+6H^(+)to2Mn^(2+)+"______"+3H_2O` (iii) `Cr_2O_7^(2-)+6Fe^(2+)+14H^(+)to2Cr^(3+)+"_______"+7H_2O`A. `3SO_4^(2-),SO_2^(2-),Fe^(3+)`B. `3SO_4^(2-),5CO_4^(2-),6Fe^(3+)`C. `3SO_4^(2-),SO_2,K^(+)`D. `S,SO_2,Fe^(3+)` |
| Answer» Correct Answer - B | |
| 400. |
For the first row transition metals the `E^(ɵ)` value are: `{:(underset(-1.18)(V),,underset(-0.91)(Cr),,underset(-1.18)(Mn),,underset(-0.44)(Fe),,underset(-0.28)(Co),,underset(-0.25)(Ni),underset(+0.34)(Cu)):}` Explain the irregularity in the above values. |
| Answer» The `E^(Ɵ)(M^(2+)//M)` values are not regular which can be explained from the irregular variation of ionisation enthalpies `(Delta_(i)H_(1)+Delta_(i)H_(2))` and also the sublimation enthalpies which are relatively much less for manganese and vanadium. | |